// We have an array A of integers, and an array queries of queries.
//
// For the i-th query val = queries[i][0], index = queries[i][1], we add val to A[index].  Then, the answer to the i-th query is the sum of the even values of A.
//
// (Here, the given index = queries[i][1] is a 0-based index, and each query permanently modifies the array A.)
//
// Return the answer to all queries.  Your answer array should have answer[i] as the answer to the i-th query.
//
//  
//
// Example 1:
//
//
// Input: A = [1,2,3,4], queries = [[1,0],[-3,1],[-4,0],[2,3]]
// Output: [8,6,2,4]
// Explanation: 
// At the beginning, the array is [1,2,3,4].
// After adding 1 to A[0], the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8.
// After adding -3 to A[1], the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6.
// After adding -4 to A[0], the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2.
// After adding 2 to A[3], the array is [-2,-1,3,6], and the sum of even values is -2 + 6 = 4.
//
//
//  
//
// Note:
//
//
// 	1 <= A.length <= 10000
// 	-10000 <= A[i] <= 10000
// 	1 <= queries.length <= 10000
// 	-10000 <= queries[i][0] <= 10000
// 	0 <= queries[i][1] < A.length
//
//


/**
 * @param {number[]} A
 * @param {number[][]} queries
 * @return {number[]}
 */
var sumEvenAfterQueries = function(A, queries) {
    let pre=new Array();
    
    for(let i=0;i<queries.length;i++){
        let arr=queries[i];
        let plus=arr[0];
        let index=arr[1];
        
        A[index]+=plus;
        let count=0;
        for(let j=0;j<A.length;j++){
            if(A[j]%2==0){
                count+=A[j];
            }
        }
        pre.push(count);
    }
    return pre;
};
